pat甲级 1152 Google Recruitment （20 分）

最新推荐文章于 2021-02-03 21:01:32 发布

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最新推荐文章于 2021-02-03 21:01:32 发布

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原文链接：http://www.cnblogs.com/8023spz/p/10104895.html

2004年，谷歌在硅谷发布了一个招聘谜题，要求求职者从自然常数e的连续数字中找出第一个10位素数。本文详细解析了这一挑战，并提供了解决方案，同时探讨了素数判断算法。

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In July 2004, Google posted on a giant billboard along Highway 101 in Silicon Valley (shown in the picture below) for recruitment. The content is super-simple, a URL consisting of the first 10-digit prime found in consecutive digits of the natural constant

The natural constant

Now you are asked to solve a more general problem: find the first K-digit prime in consecutive digits of any given L-digit number.

Input Specification:

Each input file contains one test case. Each case first gives in a line two positive integers: L ( $\leq 1,000) and K (< 10), which are the numbers of digits of the given number and the prime to be found, respectively. Then the L-digit number N is given in the next line.$

Output Specification:

For each test case, print in a line the first K-digit prime in consecutive digits of N. If such a number does not exist, output 404 instead. Note: the leading zeroes must also be counted as part of the K digits. For example, to find the 4-digit prime in 200236, 0023 is a solution. However the first digit 2 must not be treated as a solution 0002 since the leading zeroes are not in the original number.

Sample Input 1:

20 5
23654987725541023819

Sample Output 1:

Sample Input 2:

10 3
2468024680

Sample Output 2:

404

判断素数。
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX 1000
#define DMAX 10000
using namespace std;
typedef long long ll;
int l,k;
char s[MAX + 1];
bool ispri(int x) {
    if(x <= 1) return false;
    if(x == 2 || x == 3) return true;
    if(x % 6 != 1 && x % 6 != 5) return false;
    for(int i = 5;i * i <= x;i += 6) {
        if(x % i == 0 || x % (i + 2) == 0) return false;
    }
    return true;
}
int main() {
    scanf("%d%d",&l,&k);
    scanf("%s",s);
    char ch = 0;
    char ans[11] = "404";
    for(int i = 0;s[i + k - 1];i ++) {
        swap(s[i + k],ch);
        int d = atoi(s + i);
        if(ispri(d)) {
            strcpy(ans,s + i);
            break;
        }
        swap(s[i + k],ch);
    }
    puts(ans);
    return 0;
}